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I'm given the following system of equations: \begin{align*} (m+M)\ddot{x}+m\cos\alpha\ddot{s}&=-kx \qquad(1)\\ \ddot{s}+\cos\alpha\ddot{x}&=-g\sin\alpha\qquad(2)\end{align*}

I'm asked to eliminate $\ddot{s}$ and solve for $x(t)$.

Computing $m\cos\alpha \,(1) - (2)$ gives me \begin{align*} (m\cos^2\alpha-(M+m))\ddot{x}+kx=\frac{mg\sin(2\alpha)}{2} \end{align*}

Let $m\cos^2\alpha-(M+m)=\Omega$ and let $k/\Omega=L^2$.

Then I get $$\ddot{x}+L^2x=\frac{mg\sin(2\alpha)}{2\Omega}$$

So \begin{align*} x_C(t)&=A\cos(Lt)+B\sin(Lt) \\ & = A\cos\left(\sqrt{\frac{k}{m\cos^2\alpha-(M+m)}}t \right)+B\sin\left(\sqrt{\frac{k}{m\cos^2\alpha-(M+m)}}t \right)\end{align*}

With a short calculation I also find that $x_P(t)=\frac{mg\sin(2\alpha)}{2k}$, and therefore the general solution is $x(t)=x_C(t)+x_P(t)$.

Can someone verify whether this is correct or not please?

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    Not sure what you mean by $m\cos\alpha \,(1) - (2)$, but I think you reversed a sign on the following line. Substituting in $\ddot{s}$ from (2) to (1) gives $((M+m)-m\cos^2\alpha)$. Other than that, just sub in your answer to the original equation to verify that it is correct. – 2017-02-26
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    @zahbaz I mean $m\cos\alpha$ multiplied by equation $(1)$ minus equation $(2)$. – 2017-02-26
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    $-(m\cos^2\alpha-(M+m))\ddot{x}+kx=\dfrac{mg\sin(2\alpha)}{2}$ – 2017-02-26
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    Okay. Maybe check that. Not exactly sure what the system is, but if you shut your "pendulum" term off with $a=0$, you probably want your frequency $k/\Omega >0$ in order to have oscillatory solutions. Your homogeneous solution is correct besides the sign on $\Omega$. The particular solution is dimensionally correct and looks plausible. – 2017-02-26

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